CE 479
Wood Design Lecture Notes
JAR
Introduction:
Sizes of Structural Lumber and Use
Text Chapter 4
Design Approach:
The design of structural wood is carried on the basis of allowable stresses at service load
levels. Structural calculations are based on the standard net size of a piece of lumber.
Most structural lumber is dressed lumber.
1) Dressed Lumber: Lumber that has been surfaced to the standard net size, which is
less than the nominal size (stated)(Textbook Section 4.11 and NDS 01 Supplement).
i.e. 8 x12 member (nominal size = 8 x 12 in.) actually is 7 ½ x 11 ½ in. (Standard net
size) NDS Table 1A Sec. 3 Supplement 2001. Lumber is dressed on a planning
machine for the purpose of obtaining smooth surfaces and uniform sizes. Typically
lumber will be S4S (surfaced on 4 sides).
2) Rough Sawn: Large timbers are usually rough sawn to dimensions that are close to
standard net sizes, roughly 1/8” larger than the standard dressed size. Rough surface
is usually ordered specially for architectural purposes in smaller sizes.
3) Full Sawn: In this case a rough surface is obtained with actual size equal to the
nominal size.
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Wood Design Lecture Notes
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Wood Rating
The majority of sawn lumber is graded by visual inspection, and material graded in this
way (visually) is known as visually graded structural lumber. As the lumber comes out
of the mill, a person familiar with lumber grading rules examines each piece and assigns
and stamps a grade. There are two broad size classifications of sawn lumber:
•
•
Dimension Lumber: smaller (thinner) sizes of structural lumber.
Dimension lumber usually ranges in the size from 2x2 through 4x16.
In other words, dimension lumber is any material with a thickness
(smaller dimension of a piece of wood, and width is the larger
dimension) of 2 to 4 inches.
Timbers: are the larger pieces and have a minimum nominal dimension
of 5 inches. Thus, the smallest practical size timber is a 6x6 inch.
The design properties given in the NDS supplement are based on two different sets of
ASTM Standards (Textbook Sections 4.3 and 4.4):
• In-grade procedures applied to Dimension lumber
• Clear wood procedures applied to timbers
The lumber grading rules which establish the allowable stresses for use in structural
design have been developed over the years. The relative size of the wood was used as a
guide in anticipating its use. Although most lumber is visually graded, a small % of
lumber is MACHINE STRESS’ RATED by subjecting each piece of wood to a nondistructive test. This process is highly automated. As lumber comes out of the mill, it
passes through a series of rollers. In this process, a bending load is applied about the
minor axis of the cross section, and the modulus of elasticity of each piece measured. In
addition the piece is visually inspected. The material graded using MSR is limited to a
thickness of 2” or less. MSR has less variability in mechanical properties than visually
graded lumber. Consequently, is often used to fabricate engineered wood products.
•
•
Glulam beams
Wood I joists and light frame
However, stress rated boards are not commonly used for structural framing because they
are very thin. So we will focus on dimension lumber. It must be remarked that the
allowable stress depends on the species and on the size of the member.
Species (Sec. 4.5 Textbook):
A large number of species can be used to produce structural lumber. The 2001 NDS
supplement (Sec. 4, Page 29) contains allowable stresses for a large number of species.
The choice of species for use in design is a matter of economics typically. For a given
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CE 479
Wood Design Lecture Notes
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location only a few species groups may be available and it is prudent to check with local
distributors as well as a wood products agency. The species of tress used for structural
lumber are classified as hardwoods and softwoods owing not necessarily to a description
of the wood properties. For example evergreens aka conifers are a large majority of the
structural lumber. This will be either Douglas-Fir or Southern Pine.
Allowable Stresses/Design Values (NDS Tabulated values in the NDS Supplement
01): Are determined by multiplying the tabulated (stresses) by the appropriate
adjustment factors (Textbook, Sections 4.13-4.22, and design example in Section 4.23).
Thus becoming allowable design value (F’). For example for tension parallel to the grain:
Ft' = Ft x (adjustment factors) = Design value
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For an acceptable design, the axial tensile stress due to loads, ft, should not exceed the
allowable (adjusted) stress:
f t ≤ Ft'
Design Value
(stress)
Bending
Tension parallel to grain
Shear parallel to grain
Compression perpendicular
to grain
Compression parallel to
grain
Modulus of elasticity
Tabulated Stress
Fb
Ft
Pu
Fc1
Fb’
Ft’
Fv’
FcI’
Fc
Fc’
E
E’
Allowable (adjusted)
Stress
Adjustment Factors: Some decrease other increase tabulated value (Textbook, Sections
4.13-4.22, and design example in Section 4.23, and NDS 2001 Sections 2 and 4)
Examples:
C D = load duration factor
C M = web service factor (moisture content)
C F = size factor
C fu = flat use factor
C f = form factor
Stresses and adjustment factors:
Stresses due to known loads :(NDS 2001, Section 3))
f =
t
P
M
,f =
, etc.
A
S
b
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CE 479
Wood Design Lecture Notes
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Tabulated Values (Stresses): Tabulated design values listed in the NDS Supplement
2001 ED. These values include reduction for safety (F) and are for normal load duration
under the specified moisture service condition. Modulus of elasticity (E) does not
include reduction for safety and represent average values.
Dimension Lumber Page 29 NDS Supp. 2001
Table 4A, page 30, 31: Base design value for visually graded dimension lumber
(except southern pine)
Table 4B, page 36,37: Base design value for visually graded southern pine
Table 4C, page 39-42: Design values for mechanically graded dimension lumber
(MSR)
Timbers (5x5 and larger)
Table 4D, page 43-49: Design values for visually graded timbers (all species)
Adjustment Factors (Sec 4.3 NDS 01 and Suplement to NDS Tables):
A. Wet Serviced Factor: CM
EMC = Equilibrium moisture content = the average moisture content that lumber
assumes in service.
Moisture designation in grade stamp
S-Grn (surface green) MC = 19% (in service)
S-Dry (surfaced dry) MC = 15% (in service)
These values can vary depending on environmental conditions (in most buildings ranges
from 7-14% EMC). Special conditions must be analyzed individually.
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Tabulated values in NDS supplement apply to members with EMC of 19% or less
(regardless of S-GRN or S-Dry). If EMC exceeds 19%for an extended period of time,
table values should be multiplied by CM (see Page 30 and others for values in Table 4
NDS-Supp 01)
B. Load Duration Factor: CD
Wood can handle higher stresses if loads are applied for a short period of time. All
tabulated values apply to normal duration loading (10 years) The term “duration of load”
refers to the total accumulated length of time that a load is applied during the life of a
structure.
Table 2.3.2 in NDS 01 provides CD to be used in the one associated with the shortestduration of time. Whichever combination of loads, together with the appropriate load
duration factor produces the largest member size is the one that must be used in design.
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C. Size Factor: CF
The size of the member has an effect on its unit stress.
-
See Supplement 01 Tables: 4A, 4B, 4C, 4D and 4E
D. Repetitive Member Factor: Cr only “Fb”!!
The system performance of a series of small closely spaced wood members, where failure
of one member is not fatal (see Supplement 01)
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E. Flat use Factor: Cfu
Except for decking, tabulated stress for dimension lumber apply to wood members that
are stressed in flexure about the strong axis – “edgewise or load applied to narrow face”.
If however load is a applied to the wide face – the stresses may be increased by Cfu.
Tabulated bending stresses also for timber Beams & stringers apply for bending about xaxis. NDS does not provide Cfu for these cases.
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F. Temperature Factor: Ct
The strength of the wood in service is increased as the temperature cools below the
normal temp in most buildings. On the other hand, the strength decreases as temperatures
are increased. The factor Ct is the multiplier that is used to reduce tabulated stresses if
higher than normal temperatures are encountered in a design situation. Values of Ct are
given in NDS Sec. 2.3.4 for T > 100oF. Important to note that strength will be regained
when temperature returns to normal values! Thus this factor applies for sustained
conditions.
G. Form Factor: Cf
The purpose of this factor is to adjust tabulated bending stress Fb for non-rectangular
sections (see Section 3.3.4 in NDS 01).
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Example:
Determine the tabulated and allowable design values for the following member and
loading condition.
•
•
No. 2 Hem-Fir (bending about strong axis)
Floor beams 4x6 in @ 4’ on centers. Loads are (D+L). High-humidity conditions
exist, and moisture content may exceed 19%.
Stresses
Bending (NDS Supp 01)
Tabulated value, Fb = 850 psi (Tab. 4A Supp. NDS 01)
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Factors (NDS 01 Sec. 4.3)
(Table 4.3.1, NDS 01 Page 27)
In many practical situations, a number of adjustment factors may have a value of 1.0. A
comprehensive summary of the modification factors for wood members is given in NDS
Table 4.3.1
C D = load duration factors (Sec. 2.3.2 NDS)
C D = 1.0 (Table 2.3.2 controlled by live load)
C M = Wet service (Sec. 4.3.3 NDS 01, Supp Ch 4, Table 4A)
C M = 0.85 since M C > 19% or 1.0 if Fb C F ≤ 1150 psi
C F = size factor (Sec. 4.3, NDS 01 and Table 4A)
C F = 1.3
since Fb xC F = 850 x 1.3 = 1105 psi < 1150 psi
C M = 1.0
C t = Temperature factor (Sec. 4.3.4 NDS) T ≤ 100 o F
C t = 1.0
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Wood Design Lecture Notes
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C = Beam Stability Factor (Sec. 4.3.5 and 3.3.3 NSD 01)
L
6
= 1.5 < 2.0 ; no lateral support is required
4
(Sec. 3.3.3.2) C = 1.0 (Also 3.3.3.3 could be invoked if needed)
(Sec. 4.4.1.2) d / b =
L
C = 1.0 (Element is not loaded on its flat side)
fu
C = Incising Factor (Sec. 4.3.8) done to increase treatment penetrations
i
C = 1.0
i
C = Repetitive Member Factor (Sec. 4.3.9 NDS)
r
C = 1.0
r
C = Form factor (Sec. 4.3.10 &3.3.4)
f
∴ C = 1.0
f
Finally calculate allowable stress for bending
Fb' = 850 x 1 x ... x 1.3 x ... = 1105 psi
Tension II to Grain
FT = Tension parallel to grain (Table 4A Supp)
FT = 525 psi
Factors (NDS 01, Sec. 4.3 & Table 4.3.1)
C D = 1.0
C M = 1.0 (Table 4A, Supp Adj. Factors)
C t = 1.0
C F = 1.3
C i = 1.0
FT' = 525 x 1.3 = 683 psi
Shear II to Grain, FV
FV = 150 psi
Factors (NDS 4.3)
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Wood Design Lecture Notes
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C D = 1.0
C M = 0.97 Table 4A Supp Adj. Factor
C t = 1.0
C i = 1.0
FV' = 150 psi x 0.97 = 146 psi
Compression ⊥to grain
FC ⊥ = 405 psi
Factors (NDS 4.3) Table 4.3.1
(Sec. 4.3.3) and Table 4A
C M = 0.67
C t = 1.0
C i = 1.0
C b = Bearing area factor (Sec. 4.3.13)
Assume l b ≥ 6"
C b = 1.0
FC⊥ = 405 x 0.67 = 271 psi
Compression II to grain
Fc = 1300 psi
Factors (NDS 4.3 @ Table 4.3.1)
C D = 1.0 (Table 4A, Supp)
C M = 0.8 or 1.0 when (Fc ) (C F ) ≤ 750 psi
C F = 1.1 (Table 4A, Supp)
∴ C M = 0.8 since 1300 x 1.1 > 750
C t = 1 .0
C i = 1 .0
C p = 1.0 (This is a beam!)
Fc' = 1300 x 0.8 x 1.1 x K 1.0 = 1144 psi
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CE 479
Wood Design Lecture Notes
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Modulus of Elasticity, MOE
From Table 4A
E = 1,300 ,000 psi
C = 0.9 (Factors in Table 4A, Supp)
M
C = 1.0
t
C = 1.0
i
C = Buckling Stiffness factor for wood tresses (4.4.2)
T
N.A.
∴
E' = 1,300 ,000 x 0.9 = 1,170 ,000 psi
C = temperature factor
t
C = repetitive factor
r
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CE 479
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Design Summary – Beams (Chapter 6 Text)
1. Determine trial beam size based on bending stress considerations (long. Bending
stress, II to grain – see Fig. 6.1a). For sawn lumber loaded-edgewise only are given
tabulated values.
( S ) reqd =
M
Fb'
Select trial member with (use Table for dressed S4S)
(S ) prov ≥ (S )reqd
recheck for appropriate size factor, CF, since initially is unknown (beam size) so that
fb =
M
≤ Fb' (with actual C F )
(S act )
2. Check shear (Sec. 3.4 NDS)
f v = 1.5
fv =
V
≤ Fv' supp. with app. factors
A
V
td w
In this calculation a reduced shear (d- away from support face, d = overall depth) can be
used V’ (Sec. 3.4.31)(a)
f v' = 1.5
V'
A
If this check shows the beam size selected to be inadequate, the size is revised to provide
sufficient A.
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CE 479
Wood Design Lecture Notes
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Deflection Criteria (IBC 2003 Sec. 3.5 NDS 01)
Limits are established for deflections for beams, trusses, and similar members that are not
to be exceeded under certain gravity loads. Table 1604.3 in the IBC 2003 gives the
necessary limits and other information necessary to ensure user comfort and to prevent
excessive cracking of plaster ceilings.
For Green Lumber (MC > 19%)
∆ TOTAL = 2.0 (∆ long term ) + ∆ short term < L/180
∆Live < L/240
For Seasoned Lumber (MC < 19%)
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∆ TOTAL = 1.5(∆ Long Term ) + ∆ Short Term < L/180
∆Live < L/240
where
∆Long Term = immediate deflection due to the long term portion of the design load (usually
dead load)
∆Short Term = immediate deflection due to short term component of the design load (usually
live load)
Bearing - Sec. 3.10 NDS 01
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Example: Sawn Beam Design (Dimension Lumber)
•
•
•
•
Beams are spaced 16 inches on center (roof beam)
Buckling of the compression zone is prevented by the plywood roof sheathing
Material is No. 1 Douglas Fir – larch
Loads
w D = 19 lb / ft (Dead load per lineal ft.)
w L = 27 lb / ft (Live load per lineal ft.)
TOTAL = 46 lb / ft
Required load combination (Sect. 2.3.2 NDS 01 and Table 2.3.2) and Duration Factors
Dalone ⇒ C D = 0.9
D + L = C D = 1.15 ( snow load )
Determine trial size based on bending and then check other criteria.
(Sec. 4.3.9, NDS 01 and Table 4A of the supplement, Spacing < 24” ) C r = 1.15
(Table 4A Supp to NDS01) C F = 1.20
(MC < 19%, normal temperature conditions, compression edge of bending member
supported throughout in accordance with 4.4.1.2 and no incision)
C M , CT , C L and C i = 1.0
Fb = 1000 (1.15) (1.2) (1.15) = 1587 psi
Reqd S =
M max
Fb'
=
12,515
= 7.9 in 3
1587
Try 2x6 S = 7.56 in3 (From Table 1B Supp).
From NDS Supplement Table 4A
C F = 1.3
S reqd =
Fb' = 1587 x
1.3
= 1719 psi
1.2
1255
= 7.32 in 3 < 7.56 ∴ o.k .
1719
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Check for Shear (NDS Sec. 3.4)
f =
v
3V
(Rect.)
2b
d
C , C = 1.0 A = 8.25 in (Table 1B Supp)
2
M
t
Conservative to use V = V MAX = 46.0 (13.5) = 3110
fv =
3 (311)
= 56.5 psi
2 (8.25)
FV' = FV (C D ) (C M ) (C t ) (C i )
= 180 (1.15) = 207 psi > 56.5 psi ∴ ok
Check Deflections
E' = E ( C ) ( C ) ( C )( C ) = E
M
T
i
t
= 1,7000 Ksi (Table 4A Supp NDS01)
5w L
5( 27.0 )( 13.5 ) ( 1728 )
=
= 0.57"
384 E' I ( 384 ) ( 1,700 ,000 )( 20.8 )
L
13.5 x 12
∆ =
=
= 0.67" > 0.57" ∴ ok
240
240
∆ =
4
4
L
L
L
Also check for long term effects by calculating dead plus live deflection does not exceed
L/180 (Do in-class).
Use 2x6 No. 1 DF-L MC ≤ 19%
Bearing Stress Check (Sec. 3.10 NDS01)
FC' ⊥ = FC⊥ (C M ) (C t ) (C b ) = 625 psi
R
311
Re qd . ∆ =
=
= 0.49 in 2
'
625
FC
⊥
Re qd . l b >
A 0.99
=
= 0.33 : < 6" ∴ ok
b 1.5
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